Is $H^1(M) \subset L^2(M) \subset H^{-1}(M)$ a Hilbert triple for $M$ a manifold with boundary?

Question

Is $H^1(M) \subset L^2(M) \subset H^{-1}(M)$ a Hilbert triple for $M$ a manifold with boundary? What smoothness is required of the boundary? I would be grateful for some references to this.

Answer 1

I think a sufficient condition is that (it is highly probable this is not necessary): Whenever the smoothness of $\partial M$ supports a Poincaré type inequality: $$ \|u\|_{L^2(M)}\leq c\|\nabla u\|_{L^2(M)}, $$ then the Hilbert triple $$ H^1(M)/\mathbb{R}\subset L^2(M) \subset H^{-1}(M),\quad \text{ or }\quad H^1_0(M)\subset L^2_0(M) \subset H^{-1}(M) $$ holds.

The first inclusion is dense due to the fact that $C^{\infty}$ is dense in $L^2$.

Second inclusion holds if Poincaré inequality on this manifold holds for $H^1(M)$: for $\phi\in L^2(M)$ $$ L_{\phi}(u) := \int_M \phi u \leq \|\phi\|_{L^2(M)}\|u\|_{L^2(M)} \leq c\|\phi\|_{L^2(M)}\|\nabla u\|_{L^2(M)}, $$ thus $\phi\in H^{-1}(M)$.

I checked Evans's PDE book, the proof of Poincaré inequality relies on Rellich–Kondrachov compact embedding theorem, which normally assumes the boundary is Lipschitz. In Sobolev met Poincaré, the assumption is weakened to twisted cone condition for a domain (see 9.2), I am guessing this condition should be able to be extended to a compact manifold.